\(A=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\dfrac{1}{64}\)

\(=\dfrac{5+9+1}{15}-\dfrac{27+8+1}{36}+\dfrac{1}{64}\)

=1/64

a) \(A=\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{72}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)

\(=\dfrac{1}{3}-\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{1}{72}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)

\(=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\dfrac{1}{72}\)

\(=\left(\dfrac{5}{15}+\dfrac{9}{15}+\dfrac{1}{15}\right)-\left(\dfrac{27}{36}+\dfrac{8}{36}+\dfrac{1}{36}\right)+\dfrac{1}{72}\)

\(=1-1+\dfrac{1}{72}\)

\(=0+\dfrac{1}{72}=\dfrac{1}{72}\)

b) \(B=\dfrac{1}{5}-\dfrac{3}{7}+\dfrac{5}{9}-\dfrac{2}{9}+\dfrac{7}{13}-\dfrac{2}{11}-\dfrac{5}{9}+\dfrac{3}{7}-\dfrac{1}{5}\)

\(=\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\left(-\dfrac{3}{7}+\dfrac{3}{7}\right)+\left(\dfrac{5}{9}-\dfrac{5}{9}\right)-\left(\dfrac{2}{9}-\dfrac{7}{13}+\dfrac{2}{11}\right)\)

\(=0+0+0-\left(\dfrac{286}{1287}-\dfrac{693}{1287}+\dfrac{234}{1287}\right)\)

\(=-\left(-\dfrac{173}{1287}\right)\)

\(=\dfrac{173}{1287}\)

c) \(C=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-.....-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(=\dfrac{1}{100}-\left(\dfrac{1}{100.99}+\dfrac{1}{99.98}+\dfrac{1}{98.97}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\)

\(=\dfrac{1}{100}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)

\(=\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=\dfrac{1}{100}-\left(1-\dfrac{1}{100}\right)\)

\(=\dfrac{-49}{50}\)

Chắc ko

nhìn thấy nản

Thấy mà hết muốn làm

A= 3/5.6/7+3/5.3/7-3/5.2/7

A= 3/5. (6/7+3/7-2/7) = 3/5

B= (-13.2/5+-2/9.2/5+2/5.11/9).5/2

B= (-13-2/9+11/9).2/5.5/2

B= -13+(11/9-2/9)= -12

C= (-4/5+5/7).3/2+(-1/5+2/7).3/2

C= (-4/5+5/7+-1/5+2/7).3/2

C= ((-4/5+-1/5)+(5/7+2/7)).3/2

C= 0

D= 4/9.-5/3+4/9.-22/3

D= 4/9.110/9

D= 440/81

.

a) Ta có: \(\dfrac{-3}{7}+\dfrac{15}{26}-\left(\dfrac{2}{13}-\dfrac{3}{7}\right)\)

\(=\dfrac{-3}{7}+\dfrac{15}{26}-\dfrac{2}{13}+\dfrac{3}{7}\)

\(=\dfrac{15}{26}-\dfrac{4}{26}\)

\(=\dfrac{11}{26}\)

b) Ta có: \(2\cdot\dfrac{3}{7}+\left(\dfrac{2}{9}-1\dfrac{3}{7}\right)-\dfrac{5}{3}:\dfrac{1}{9}\)

\(=\dfrac{6}{7}+\dfrac{2}{9}-\dfrac{10}{7}-\dfrac{5}{3}\cdot9\)

\(=\dfrac{-4}{7}+\dfrac{2}{9}-15\)

\(=\dfrac{-36}{63}+\dfrac{14}{63}-\dfrac{945}{63}\)

\(=\dfrac{-967}{63}\)

c) Ta có: \(\dfrac{-11}{23}\cdot\dfrac{6}{7}+\dfrac{8}{7}\cdot\dfrac{-11}{23}-\dfrac{1}{23}\)

\(=\dfrac{-11}{23}\cdot\left(\dfrac{6}{7}+\dfrac{8}{7}\right)-\dfrac{1}{23}\)

\(=\dfrac{-11}{23}\cdot2-\dfrac{1}{23}\)

\(=-1\)

d) Ta có: \(\left(\dfrac{377}{-231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot\left(\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{24}\right)\)

\(=\left(\dfrac{-377}{231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot\left(\dfrac{4}{24}-\dfrac{3}{24}-\dfrac{1}{24}\right)\)

\(=\left(\dfrac{-377}{231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot0\)

=0

A -\(\dfrac{24}{25}\)

B -\(\dfrac{5}{21}\)

C -\(\dfrac{24}{47}\)

D -\(\dfrac{19}{42}\)

tick cho mk

a: =11+3/4-6-5/6+4+1/2+1+2/3

=10+9/12-10/12+6/12+8/12

=10+13/12=133/12

b: \(=2+\dfrac{17}{20}-1-\dfrac{11}{15}+2+\dfrac{3}{20}\)

=3-11/15

=34/15

c: \(=\dfrac{31}{7}:\left(\dfrac{7}{5}\cdot\dfrac{31}{7}\right)\)

\(=\dfrac{31}{7}:\dfrac{31}{5}=\dfrac{5}{7}\)

d: \(=\dfrac{29}{8}\cdot\dfrac{36}{29}\cdot\dfrac{15}{23}\cdot\dfrac{23}{5}=\dfrac{9}{2}\cdot3=\dfrac{27}{2}\)

a,\(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}+\dfrac{11}{13}-\dfrac{9}{11}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}\)

\(=\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(-\dfrac{3}{5}+\dfrac{3}{5}\right)+.....+\left(-\dfrac{11}{13}+\dfrac{11}{13}\right)+\dfrac{13}{15}\)

\(=0+0+...0+0+\dfrac{13}{15}=\dfrac{13}{15}\)

câu b và c xem lại đề nha

Chúc bạn học tốt!!!

b, \(\dfrac{1}{99}+\dfrac{1}{99.98}+\dfrac{1}{98.97}+......+\dfrac{1}{3.2}+\dfrac{1}{2.1}\)

\(=\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{97}+.........+\dfrac{1}{3}-\dfrac{1}{2}+\dfrac{1}{2}-1\)

(do \(\dfrac{n}{a.\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\))

\(=\dfrac{1}{99}+\dfrac{1}{99}-1=\dfrac{2}{99}-1=\dfrac{-97}{99}\)

Chúc bạn học tốt!!!

c/ \(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}+...+\dfrac{1}{155\cdot257}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}+\dfrac{2}{255\cdot257}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}+...+\dfrac{1}{255}-\dfrac{1}{257}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{257}\right)=\dfrac{1}{2}\cdot\dfrac{256}{257}=\dfrac{128}{257}\)

p/s: thấy bn kia hơn 20' r` k lm tiếp nên t lm lun nhé!